This is a 'finite calculus' (differences & summations) problem.
You can solve it using difference operator (actually its inverse which gives
you the discrete integration which is nothing but summation).
If you do not know finite calculus, Google for it (or refer Concrete
Mathematics by Knuth).
The solution for any k is.
*f(n) = nC(k+1) + nC(k-1) + nC(k-3) + .... (all the way down to nC0 or nC1
depends on k is odd or even).*
Here nCr is the binomial coefficient "n choose r".
Eg: Let k = 3, n = 4
f(4) = 4C4 + 4C2 + 4C0 = 1 + 6 + 1 = 8
Another, k = 3 and n = 5
f(5) = 5C4 + 5C2 + 5C0 = 5 + 10 + 1 = 16
On Wed, Nov 4, 2009 at 11:23 AM, abhijith reddy <abhijith200...@gmail.com>wrote:
> Is there a way to find the sum of the Kth series ( Given below)
> K=0 S={1,2,3,4,5,6,....}
> K=1 S={1,2,4,7,11,16..} common diff = 1,2,3,4 5 ...
> K=2 S={1,2,4,8,15,26...} common diff = 1,2,4,7 11... (series with
> K=1)
> K=3 S={1,2,4,8,16,31...} common diff = 1,2,4,8 15... (series with
> K=2)
> Note that the common difference of Kth series is the (K-1) series
> Any ideas ??
> --
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On a related note,
The solution I gave you is to find the nth element in the kth series.
If you want to sum the first 'n' elements of the kth series (call the
function s(n,k)), then it is easy to see that,
*s(n,k) = f(n+1, k+1) - 1*
where f(n+1, k+1) is the (n+1)th element in the (k+1)th series.
This can also be easily done using the summation operator of 'finite
calculus'.
On Fri, Nov 6, 2009 at 10:50 AM, Prunthaban Kanthakumar <
pruntha...@gmail.com> wrote:
> This is a 'finite calculus' (differences & summations) problem.
> You can solve it using difference operator (actually its inverse which
> gives you the discrete integration which is nothing but summation).
> If you do not know finite calculus, Google for it (or refer Concrete
> Mathematics by Knuth).
> The solution for any k is.
> *f(n) = nC(k+1) + nC(k-1) + nC(k-3) + .... (all the way down to nC0 or nC1
> depends on k is odd or even).*
> Here nCr is the binomial coefficient "n choose r".
> Eg: Let k = 3, n = 4
> f(4) = 4C4 + 4C2 + 4C0 = 1 + 6 + 1 = 8
> Another, k = 3 and n = 5
> f(5) = 5C4 + 5C2 + 5C0 = 5 + 10 + 1 = 16
> On Wed, Nov 4, 2009 at 11:23 AM, abhijith reddy <abhijith200...@gmail.com>wrote:
>> Is there a way to find the sum of the Kth series ( Given below)
>> K=0 S={1,2,3,4,5,6,....}
>> K=1 S={1,2,4,7,11,16..} common diff = 1,2,3,4 5 ...
>> K=2 S={1,2,4,8,15,26...} common diff = 1,2,4,7 11... (series with
>> K=1)
>> K=3 S={1,2,4,8,16,31...} common diff = 1,2,4,8 15... (series with
>> K=2)
>> Note that the common difference of Kth series is the (K-1) series
>> Any ideas ??
>> --
>> You received this message because you are subscribed to the Google Groups
>> "Algorithm Geeks" group.
>> To post to this group, send email to algogeeks@googlegroups.com.
>> For more options, visit this group at
>> http://groups.google.com/group/algogeeks?hl=en.
pruntha...@gmail.com> wrote:
> On a related note,
> The solution I gave you is to find the nth element in the kth series.
> If you want to sum the first 'n' elements of the kth series (call the
> function s(n,k)), then it is easy to see that,
> *s(n,k) = f(n+1, k+1) - 1*
> where f(n+1, k+1) is the (n+1)th element in the (k+1)th series.
> This can also be easily done using the summation operator of 'finite
> calculus'.
> On Fri, Nov 6, 2009 at 10:50 AM, Prunthaban Kanthakumar <
> pruntha...@gmail.com> wrote:
>> This is a 'finite calculus' (differences & summations) problem.
>> You can solve it using difference operator (actually its inverse which
>> gives you the discrete integration which is nothing but summation).
>> If you do not know finite calculus, Google for it (or refer Concrete
>> Mathematics by Knuth).
>> The solution for any k is.
>> *f(n) = nC(k+1) + nC(k-1) + nC(k-3) + .... (all the way down to nC0 or
>> nC1 depends on k is odd or even).*
>> Here nCr is the binomial coefficient "n choose r".
>> Eg: Let k = 3, n = 4
>> f(4) = 4C4 + 4C2 + 4C0 = 1 + 6 + 1 = 8
>> Another, k = 3 and n = 5
>> f(5) = 5C4 + 5C2 + 5C0 = 5 + 10 + 1 = 16
>> On Wed, Nov 4, 2009 at 11:23 AM, abhijith reddy <abhijith200...@gmail.com
>> > wrote:
>>> Is there a way to find the sum of the Kth series ( Given below)
>>> K=0 S={1,2,3,4,5,6,....}
>>> K=1 S={1,2,4,7,11,16..} common diff = 1,2,3,4 5 ...
>>> K=2 S={1,2,4,8,15,26...} common diff = 1,2,4,7 11... (series with
>>> K=1)
>>> K=3 S={1,2,4,8,16,31...} common diff = 1,2,4,8 15... (series with
>>> K=2)
>>> Note that the common difference of Kth series is the (K-1) series
>>> Any ideas ??
>>> --
>>> You received this message because you are subscribed to the Google Groups
>>> "Algorithm Geeks" group.
>>> To post to this group, send email to algogeeks@googlegroups.com.
>>> For more options, visit this group at
>>> http://groups.google.com/group/algogeeks?hl=en.
> --
> You received this message because you are subscribed to the Google Groups
> "Algorithm Geeks" group.
> To post to this group, send email to algogeeks@googlegroups.com.
> For more options, visit this group at
> http://groups.google.com/group/algogeeks?hl=en.
> On Fri, Nov 6, 2009 at 11:00 AM, Prunthaban Kanthakumar <
> pruntha...@gmail.com> wrote:
>> On a related note,
>> The solution I gave you is to find the nth element in the kth series.
>> If you want to sum the first 'n' elements of the kth series (call the
>> function s(n,k)), then it is easy to see that,
>> *s(n,k) = f(n+1, k+1) - 1*
>> where f(n+1, k+1) is the (n+1)th element in the (k+1)th series.
>> This can also be easily done using the summation operator of 'finite
>> calculus'.
>> On Fri, Nov 6, 2009 at 10:50 AM, Prunthaban Kanthakumar <
>> pruntha...@gmail.com> wrote:
>>> This is a 'finite calculus' (differences & summations) problem.
>>> You can solve it using difference operator (actually its inverse which
>>> gives you the discrete integration which is nothing but summation).
>>> If you do not know finite calculus, Google for it (or refer Concrete
>>> Mathematics by Knuth).
>>> The solution for any k is.
>>> *f(n) = nC(k+1) + nC(k-1) + nC(k-3) + .... (all the way down to nC0 or
>>> nC1 depends on k is odd or even).*
>>> Here nCr is the binomial coefficient "n choose r".
>>> Eg: Let k = 3, n = 4
>>> f(4) = 4C4 + 4C2 + 4C0 = 1 + 6 + 1 = 8
>>> Another, k = 3 and n = 5
>>> f(5) = 5C4 + 5C2 + 5C0 = 5 + 10 + 1 = 16
>>> On Wed, Nov 4, 2009 at 11:23 AM, abhijith reddy <
>>> abhijith200...@gmail.com> wrote:
>>>> Is there a way to find the sum of the Kth series ( Given below)
>>>> K=0 S={1,2,3,4,5,6,....}
>>>> K=1 S={1,2,4,7,11,16..} common diff = 1,2,3,4 5 ...
>>>> K=2 S={1,2,4,8,15,26...} common diff = 1,2,4,7 11... (series with
>>>> K=1)
>>>> K=3 S={1,2,4,8,16,31...} common diff = 1,2,4,8 15... (series with
>>>> K=2)
>>>> Note that the common difference of Kth series is the (K-1) series
>>>> Any ideas ??
>>>> --
>>>> You received this message because you are subscribed to the Google
>>>> Groups "Algorithm Geeks" group.
>>>> To post to this group, send email to algogeeks@googlegroups.com.
>>>> For more options, visit this group at
>>>> http://groups.google.com/group/algogeeks?hl=en.
>> --
>> You received this message because you are subscribed to the Google Groups
>> "Algorithm Geeks" group.
>> To post to this group, send email to algogeeks@googlegroups.com.
>> For more options, visit this group at
>> http://groups.google.com/group/algogeeks?hl=en.
> --
> You received this message because you are subscribed to the Google Groups
> "Algorithm Geeks" group.
> To post to this group, send email to algogeeks@googlegroups.com.
> For more options, visit this group at
> http://groups.google.com/group/algogeeks?hl=en.
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